We have \(2^x = y^2-1 = (y-1)(y+1)\). Since the factors \((y-1)\) and \((y+1)\) on the right hand side are integers whose product is a power of 2, both \((y-1)\) and \((y+1)\) must be powers of 2. Furthermore, their difference is

\[ (y+1)-(y-1)=2,\]

implying the factors must be \(y+1 = 4\) and \(y-1 = 2\). This gives \( y=3\), and thus \(x=3\). Therefore, \((3, 3)\) is the only solution. \(_\square\)

Factorize the polynomial

\[f(a, b, c) = ab(a^2-b^2) + bc(b^2-c^2) + ca(c^2-a^2).\]

Observe that if \( a=b\), then \(f(a, a, c) =0\); if \(b=c\), then \(f(a, b, b)=0\); and if \( c=a\), then \( f(c,b,c)=0\). By the Remainder-Factor Theorem, \( (a-b), (b-c),\) and \( (c-a)\) are factors of \( f(a,b,c)\). This allows us to factorize